php CodeIgniter - 在视图中链接到另一个页面的正确方法
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CodeIgniter - Correct way to link to another page in a view
提问by Cecil
I was wondering if someone could tell me the correct way to link to another page from within a view.
我想知道是否有人可以告诉我从视图中链接到另一个页面的正确方法。
Is there a function for this or is it just the usual about
是否有此功能或只是通常的
Cheers,
干杯，
回答by jondavidjohn
I assume you are meaning "internally" within your application.
我假设您在应用程序中的意思是“内部”。
you can create your own <a>tag and insert a url in the href like this
您可以创建自己的<a>标签并像这样在 href 中插入一个 url
<a href="<?php echo site_url('controller/function/uri') ?>">Link</a>
OR you can use the URL helper this way to generate an <a>tag
或者您可以通过这种方式使用 URL 助手来生成<a>标签
anchor(uri segments, text, attributes)
So... to use it...
所以……使用它……
<?php echo anchor('controller/function/uri', 'Link', 'class="link-class"') ?>
and that will generate
这将产生
<a href="http://domain.com/index.php/controller/function/uri" class="link-class">Link</a>
For the additional commented question
对于额外的评论问题
I would use my first example
我会用我的第一个例子
so...
所以...
<a href="<?php echo site_url('controller/function') ?>"><img src="<?php echo base_url() ?>img/path/file.jpg" /></a>
for images (and other assets) I wouldn't put the file path within the php, I would just echo the base_url() and then add the path normally.
对于图像（和其他资产），我不会将文件路径放在 php 中，我只会回显 base_url() 然后正常添加路径。
回答by Ashish Bafna
The best way is to use the following code:
最好的方法是使用以下代码：
<a href="<?php echo base_url() ?>directory_name/filename.php">Link</a>
回答by winnie damayo
you can also use PHP short tag to make it shorter. here's an example
你也可以使用 PHP 短标签来缩短它。这是一个例子
<a href="<?= site_url('controller/function'); ?>Contacts</a>
or use the built in anchor function of CI.
或使用 CI 的内置锚点功能。
回答by Ghulam Abbas
you can also use this code
您也可以使用此代码
//test" class="btn btn-primary pull-right">
//test" class="btn btn-primary pull-right">
回答by Amaan Iqbal
Best and easiest way is to use anchor tag in CodeIgniter like eg.
最好和最简单的方法是在 CodeIgniter 中使用锚标记，例如。
<?php
$this->load->helper('url');
echo anchor('name_of_controller_file/function_name_if_any', 'Sign Out', array('class' => '', 'id' => ''));
?>
Refer https://www.codeigniter.com/user_guide/helpers/url_helper.htmlfor details
请参阅https://www.codeigniter.com/user_guide/helpers/url_helper.html的细节
This will surely work.
这肯定会奏效。
回答by jareenamol joy
<a href="<?php echo site_url('controller/function'); ?>Compose</a>
<a href="<?php echo site_url('controller/function'); ?>Inbox</a>
<a href="<?php echo site_url('controller/function'); ?>Outbox</a>
<a href="<?php echo site_url('controller/function'); ?>logout</a>
<a href="<?php echo site_url('controller/function'); ?>logout</a>
